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I think there is no need to sort complete array, we can make use of min-heap here. Complete algorithm can complete in O(n/2 * log n)
Steps:
1. Transform array into min-heap, in O(n)
2. Swap root with second last element in array
3. Reduce heap size by 2 (which means last 2 elements are already in required form)
4. Heapify root element.
5. Repeat steps 3 and 4 until heap size becomes 0 (this operation completes in O(n/2 * log n))
Please refer below source code
- IntMainReturnZero February 25, 2014